NYOJ Cut the rope

Cut the rope

时间限制： 1000 ms  |  内存限制： 65535 KB

难度： 4

描述

 

We have a rope whose length is L. We will cut the rope into two or more parts, the length of each part must be an integer, and no two parts have the same length.
　　Your task is to calculate there exists how many results after the cutting. We say two results are different if and only if there is at least one part, which we can find in one result but can not find in the other result

输入

There is an integer T (1 <= T <= 50000) in the first line, which indicates there are T test cases in total.
　　For each test case, there is only one integer L (1 <= L <= 50000) which has the same meaning as above.

输出

For each test case, you should output the correct answer of the above task in one line.
　　Because the answer may be very large, you should just output the remainder of it divided by 1000000 instead

样例输入

3
2
3
6

样例输出

0
1
3

上传者

动态规划，dp[i][j]表示i分解陈j个数的数量,

dp[i][j]=dp[i-j][j]（不包含1）+dp[i-j][j-1]（包含1）;

AC代码：

# include <cstdio>
# include <cstring>
# include <cmath>
# include <cstdlib>
using namespace std;
const int mod=1000000;
int dp[50010][320];
int main(){
	int n, t, i, j, k, sum;
	memset(dp, 0, sizeof(dp));
	dp[3][2]=1;dp[4][2]=1;
	dp[5][2]=2;dp[6][2]=2;dp[6][3]=1;
	for(i=1; i<=50000; i++){
		dp[i][1]=1;
	}
	for(i=7; i<=50000; i++){
		for(j=2; j*(j+1)/2<=i; j++){
			dp[i][j]=(dp[i-j][j]+dp[i-j][j-1])%mod;
		}
	}
	scanf("%d", &t);
	for(i=1; i<=t; i++){
		scanf("%d", &n);
		sum=0;
		for(k=2; k*(k+1)/2<=n; k++){
			sum=(sum+dp[n][k])%mod;
		}
		printf("%d\n", sum);
	}
	return 0;
}